Find the derivative of the given function.
$f(t)=\(\csc\)^{-1}(2t+7)$

• A. $-\(\frac{2}{\left|t\right|\sqrt{t^2-1}\)}$
• B. $-\(\frac{2}{\left|t\right|\sqrt{\left(2t+7\right)^2-1}\)}$
• C. $-\(\frac{2t}{\left|2t+7\right|\sqrt{\left(2t+7\right)^2-1}\)}$
• D. $-\(\frac{2}{\left|2t+7\right|\sqrt{\left(2t+7\right)^2-1}\)}$

Answer: (D)

Verified step by step guidance

1. Identify the function for which you need to find the derivative: \( f(t) = \csc^{-1}(2t+7) \).
2. Recall the derivative formula for the inverse cosecant function: \( \frac{d}{dx}[\csc^{-1}(x)] = -\frac{1}{|x|\sqrt{x^2-1}} \).
3. Apply the chain rule to differentiate \( f(t) = \csc^{-1}(2t+7) \). The chain rule states that \( \frac{d}{dt}[\csc^{-1}(u)] = \frac{d}{du}[\csc^{-1}(u)] \cdot \frac{du}{dt} \), where \( u = 2t+7 \).
4. Differentiate \( u = 2t+7 \) with respect to \( t \), which gives \( \frac{du}{dt} = 2 \).
5. Substitute \( u = 2t+7 \) and \( \frac{du}{dt} = 2 \) into the chain rule expression: \( \frac{d}{dt}[\csc^{-1}(2t+7)] = -\frac{2}{|2t+7|\sqrt{(2t+7)^2-1}} \).